NCERT solutions for class 9 maths chapter 10 Circles: A round figure made up using several points that are equidistant from a reference point is called a circle. You can relate this chapter with reallife things. Many objects that we come across in our daily life is circular in shape, such as ring, bangle, wheels of the vehicle, clock, etc. CBSE NCERT solutions for class 9 maths chapter 10 Circles will help in solving all problems related to circular shapes. The circle is an integral part of unit geometry. A circle divides the plane in which the circle lies into three parts as shown in figure 1, which are the interior of the circle, the exterior of the circle and the circle. In this particular chapter, you will learn the proof of theorems and problems based on those theorems. Solutions of NCERT for class 9 maths chapter 10 Circles are designed in such a way that a student can fetch 100% marks in a particular question. In the upcoming class, you will get to know some advanced versions of circles but the advanced part can only be done when you have done the basics. So, class 9 maths circles are a foundation for next class board examinations. NCERT solutions for class 9 maths chapter 10 circles will assist you in building this strong foundation. NCERT solutions are also available for different chapters, classes, and subjects which you can get by clicking on the given link.
Fill in the blanks:
Q1 (i) The centre of a circle lies in _____________ of the circle. (exterior/ interior)
The centre of a circle lies in the interior of the circle.
Fill in the blanks:
A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.
Fill in the blanks:
Q1 (iii) The longest chord of a circle is a ___________ of the circle.
The longest chord of a circle is a diameter of the circle.
Fill in the blanks:
Q1 (iv) An arc is a ________________ when its ends are the ends of a diameter.
An arc is a semi circle when its ends are the ends of a diameter.
Fill in the blanks:
Q1 (v) Segment of a circle is the region between an arc and ________________ of the circle.
Segment of a circle is the region between an arc and chord of the circle.
Fill in the blanks:
Q1 (vi) A circle divides the plane, on which it lies, in _______________ parts.
A circle divides the plane, on which it lies, in two parts.
Write True or False: Give reasons for your answers.
Q2 (i) Line segment joining the centre to any point on the circle is a radius of the circle.
True. As line segment joining the centre to any point on the circle is a radius of the circle.
Write True or False: Give reasons for your answers.
Q2 (ii) A circle has only finite number of equal chords.
False . As a circle has infinite number of equal chords.
Write True or False: Give reasons for your answers.
Q2 (iii) If a circle is divided into three equal arcs, each is a major arc.
False. If a circle is divided into three equal arcs, each arc makes angle of 120 degrees whereas major arc makes angle greater than 180 degree at centre.
Write True or False: Give reasons for your answers.
Q2 (iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
True.A chord of a circle, which is twice as long as its radius, is a diameter of the circle.
Write True or False: Give reasons for your answers.
Q2 (v) Sector is the region between the chord and its corresponding arc.
False. As the sector is the region between the radii and arc.
Write True or False: Give reasons for your answers.
Q2 (vi) A circle is a plane figure.
Given: The two circles are congruent if they have the same radii.
To prove: The equal chords of congruent circles subtend equal angles at their centres i.e. BAC=QPR
Proof :
In ABC and PQR,
BC = QR (Given)
AB = PQ (Radii of congruent circle)
AC = PR (Radii of congruent circle)
Thus, ABC PQR (By SSS rule)
BAC=QPR (CPCT)
Given : chords of congruent circles subtend equal angles at their centres,
To prove : BC = QR
Proof :
In ABC and PQR,
BAC=QPR (Given)
AB = PQ (Radii of congruent circle)
AC = PR (Radii of congruent circle)
Thus, ABC PQR (By SAS rule)
BC = QR (CPCT)
In (i) we do not have any common point.
In (ii) we have 1 common point.
In (iii) we have 1 common point.
In (iv) we have 2 common points.
The maximum number of common points is 2.
Q2 Suppose you are given a circle. Give a construction to find its centre.
Answer:
Given : Points P,Q,R lies on circle.
Construction :
1. Join PR and QR
2. Draw perpendicular bisector of PR and QR which intersects at point O.
3. Taking O as centre and OP as radius draw a circle.
4. The circle obtained is required.
Given: Two circles intersect at two points.
To prove: their centres lie on the perpendicular bisector of the common chord.
Construction: Joinpoint P and Q to midpoint M of chord AB.
Proof: AB is a chord of circle C(Q,r) and QM is the bisector of chord AB.
Hence,
Similarly, AB is a chord of circle(Q,r' ) and QM is the bisector of chord AB.
Hence,
Now,
PMA and QMA are forming linear pairs so PMQ is a straight line.
Hence, P and Q lie on the perpendicular bisector of common chord AB.
Given: Two circles of radii and intersect at two points and the distance between their centres is .
To find the length of the common chord.
Construction: Join OP and draw
Proof: AB is a chord of circle C(P,3) and PM is the bisector of chord AB.
Let, PM = x , so QM=4x
In APM, using Pythagoras theorem
...........................1
Also,
In AQM, using Pythagoras theorem
...........................2
From 1 and 2, we get
Put,x=0 in equation 1
Given: two equal chords of a circle intersect within the circle
To prove: Segments of one chord are equal to corresponding segments of the other chord i.e. AP = CP and BP=DP.
Construction : Join OP and draw
Proof :
In OMP and ONP,
AP = AP (Common)
OM = ON (Equal chords of a circle are equidistant from the centre)
OMP = ONP (Both are right angled)
Thus, OMP ONP (By SAS rule)
PM = PN..........................1 (CPCT)
AB = CD ............................2(Given )
......................3
Adding 1 and 3, we have
AM + PM = CN + PN
Subtract 4 from 2, we get
ABAP = CD  CP
Given: two equal chords of a circle intersect within the circle.
To prove: the line joining the point of intersection to the centre makes equal angles with the chords.
i.e. OPM=OPN
Proof :
Construction: Join OP and draw
In OMP and ONP,
AP = AP (Common)
OM = ON (Equal chords of a circle are equidistant from the centre)
OMP = ONP (Both are rightangled)
Thus, OMP ONP (By RHS rule)
OPM=OPN (CPCT)
Given: a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D.
To prove : AB = CD
Construction: Draw
Proof :
BC is a chord of the inner circle and
So, BM = CM .................1
(Perpendicular OM bisect BC)
Similarly,
AD is a chord of the outer circle and
So, AM = DM .................2
(Perpendicular OM bisect AD )
Subtracting 1 from 2, we get
AMBM = DM  CM
Given: From the figure, R, S, M are the position of Reshma, Salma, Mandip respectively.
So, RS = SM = 6 cm
Construction : Join OR,OS,RS,RM and OM.Draw .
Proof:
In ORS,
OS = OR and (by construction )
So, RL = LS = 3cm (RS = 6 cm )
In OLS, by pytagoras theorem,
In ORK and OMK,
OR = OM (Radii)
ROK = MOK (Equal chords subtend equal angle at centre)
OK = OK (Common)
ORK OMK (By SAS)
RK = MK (CPCT)
Thus,
area of ORS =...............................1
area of ORS =.............................2
From 1 and 2, we get
Thus,
Given: In the figure, A, S, D are positioned Ankur, Syed and David respectively.
So, AS = SD = AD
Radius of circular park = 20 m
so, AO=SO=DO=20 m
Construction: AP SD
Proof :
Let AS = SD = AD = 2x cm
In ASD,
AS = AD and AP SD
So, SP = PD = x cm
In OPD, by Pythagoras,
In APD, by Pythagoras,
Squaring both sides,
Hence, length of string of each phonem
AOC = AOB + BOC=
AOC = 2 ADC (angle subtended by an arc at the centre is double the angle subtended by it at any)
Given: A chord of a circle is equal to the radius of the circle i.e. OA=OB.
To find: ADB and ACB.
Solution :
In OAB,
OA = AB (Given )
OA = OB (Radii of circle)
So, OA=OB=AB
ABC is a equilateral triangle.
So,AOB =
AOB = 2 ADB
ACBD is a cyclic quadrilateral .
So, ACB+ADB =
Q3 In Fig. , , where P, Q and R are points on a circle with centre O. Find .
Construction: Join PS and RS.
PQRS is a cyclic quadrilateral.
So, PSR + PQR =
Here, POR = 2 PSR
In OPR ,
OP=OR (Radii )
ORP = OPR (the angles opposite to equal sides)
In OPR ,
OPR+ORP+POR=
.
DEC+ BEC = (linear pairs)
DEC+ = ( BEC =)
DEC = 
DEC =
In DEC,
D+ DEC+ DCE =
D = BAC (angles in same segment are equal )
BAC =
(angles in the same segment are equal )
In
If AB = BC ,then
Here,
AC is the diameter of the circle.
Thus, and ............................1(Angle in a semicircle is right angle)
Similarly, BD is the diameter of the circle.
Thus, and ............................2(Angle in a semicircle is right angle)
From 1 and 2, we get
Hence, ABCD is a rectangle.
Q8 If the nonparallel sides of a trapezium are equal, prove that it is cyclic.
Given: ABCD is a trapezium.
Construction: Draw AD  BE.
Proof: In quadrilateral ABED,
AB  DE (Given )
AD  BE ( By construction )
Thus, ABED is a parallelogram.
AD = BE (Opposite sides of parallelogram )
AD = BC (Given )
so, BE = BC
In EBC,
BE = BC (Proved above )
Thus, ...........1(angles opposite to equal sides )
...............2(Opposite angles of the parallelogram )
From 1 and 2, we get
(linear pair)
Thus, ABED is a cyclic quadrilateral.
................1(vertically opposite angles)
..................2(Angles in the same segment are equal)
.................3(angles in the same segment are equal)
From 1,2,3 ,we get
Given: circles are drawn taking two sides of a triangle as diameters.
Construction: Join AD.
Proof: AB is the diameter of the circle and ADB is formed in a semicircle.
ADB = ........................1(angle in a semicircle)
Similarly,
AC is the diameter of the circle and ADC is formed in a semicircle.
ADC = ........................2(angle in a semicircle)
From 1 and 2, we have
ADB+ADC=+=
ADB and ADC are forming a linear pair. So, BDC is a straight line.
Hence, point D lies on this side.
Q11 ABC and ADC are two right triangles with common hypotenuse AC. Prove that
.
Given: ABC and ADC are two right triangles with common hypotenuse AC.
To prove :
Proof :
Triangle ABC and ADC are on common base BC and BAC = BDC.
Thus, point A,B,C,D lie in the same circle.
(If a line segment joining two points subtends equal angles at two other points lying on the same side of line containing line segment, four points lie on the circle.)
CAD = CBD (Angles in same segment are equal)
Q12 Prove that a cyclic parallelogram is a rectangle.
Answer:
Given: ABCD is a cyclic quadrilateral.
To prove: ABCD is a rectangle.
Proof :
In cyclic quadrilateral ABCD.
.......................1(sum of either pair of opposite angles of a cyclic quadrilateral)
........................................2(opposite angles of a parallelogram are equal )
From 1 and 2,
We know that a parallelogram with one angle right angle is a rectangle.
Hence, ABCD is a rectangle.
Given: Circle C(P,r) and circle C(Q,r') intersect each other at A and B.
To prove : PAQ = PBQ
Proof : In APQ and BPQ,
PA = PB (radii of same circle)
PQ = PQ (Common)
QA = QB (radii of same circle)
So, APQ BPQ (By SSS)
PAQ = PBQ (CPCT)
Given : AB = 5 cm, CD = 11 cm and AB  CD.
To find Radius (OA).
Construction: Draw
Proof :
Proof: CD is a chord of circle and
Thus, CM = MD = 5.5 cm (perpendicular from centre bisects chord)
and AN = NB = 2.5 cm
Let OM be x.
So, ON = 6  x (MN = 6 cm )
In OCM , using Pythagoras,
.............................1
and
In OAN , using Pythagoras,
.............................2
From 1 and 2,
(OC=OA =radii)
From 2, we get
OA = OC
Thus, the radius of the circle is
Given : AB = 8 cm, CD = 6 cm , OM = 4 cm and AB  CD.
To find: Length of ON
Construction: Draw
Proof :
Proof: CD is a chord of circle and
Thus, CM = MD = 3 cm (perpendicular from centre bisects chord)
and AN = NB = 4 cm
Let MN be x.
So, ON = 4  x (MN = 4 cm )
In OCM , using Pythagoras,
.............................1
and
In OAN , using Pythagoras,
.............................2
From 1 and 2,
(OC=OA =radii)
So, x=1 (since )
ON =4x =41=3 cm
Hence, second chord is 3 cm away from centre.
Given : AD = CE
To prove :
Construction: Join AC and DE.
Proof :
Let ADC = x , DOE = y and AOD = z
So, EOC = z (each chord subtends equal angle at centre)
AOC + DOE +AOD + EOC =
.........................................1
In OAD ,
OA = OD (Radii of the circle)
OAD = ODA (angles opposite to equal sides )
OAD + ODA + AOD =
.............................................................2
Similarly,
.............................................................3
..............................................................4
ODB is exterior of triangle OAD . So,
ODB = OAD + ODA
(from 2)
.................................................................5
similarly,
OBE is exterior of triangle OCE . So,
OBE = OCE + OEC
(from 3)
.................................................................6
From 4,5,6 ;we get
BDE = BED = OEB  OED
..................................................7
In BDE ,
DBE + BDE + BED =
...................................................8
Here, from equation 1,
...................................9
From 8 and 9,we have
Given : ABCD is rhombus.
To prove : the circle drawn with AB as diameter, passes through the point O.
Proof :
ABCD is rhombus.
Thus, (diagonals of a rhombus bisect each other at )
So, a circle drawn AB as diameter will pass through point O.
Thus, the circle is drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
Given: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.
To prove : AE = AD
Proof :
ADC = 3 , ABC = 4, ADE = 1 and AED = 2
.................1(linear pair)
....................2(sum of opposite angles of cyclic quadrilateral)
3 = 4 (oppsoite angles of parallelogram )
From 1 and 2,
3+1 = 2 + 4
From 3, 1 = 2
From 4, AQB, 1 = 2
Therefore, AE = AD (In an isosceles triangle ,angles oppsoite to equal sides are equal)
Q7 (i) AC and BD are chords of a circle which bisect each other. Prove that AC and BD are diameters
Given: AC and BD are chords of a circle which bisect each other.
To prove: AC and BD are diameters.
Construction : Join AB,BC,CD,DA.
Proof :
In ABD and CDO,
AO = OC (Given )
AOB = COD (Vertically opposite angles )
BO = DO (Given )
So, ABD CDO (By SAS)
BAO = DCO (CPCT)
BAO and DCO are alternate angle and are equal .
So, AB  DC ..............1
Also AD  BC ...............2
From 1 and 2,
......................3(sum of opposite angles)
A = C ................................4(Opposite angles of the parallelogram )
From 3 and 4,
BD is a diameter of the circle.
Similarly, AC is a diameter.
Q7 (ii) AC and BD are chords of a circle which bisect each other. Prove that ABCD is a rectangle.
Given: AC and BD are chords of a circle which bisect each other.
To prove: ABCD is a rectangle.
Construction : Join AB,BC,CD,DA.
Proof :
ABCD is a parallelogram. (proved in (i))
(proved in (i))
A parallelogram with one angle , is a rectangle )
Thus, ABCD is rectangle.
Given : Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.
To prove : the angles of the triangle DEF are , and
Proof :
1 and 3 are angles in same segment.therefore,
1 = 3 ................1(angles in same segment are equal )
and 2 = 4 ..................2
Adding 1 and 2,we have
1+2=3+4
,
and
Similarly, and
Given: Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles.
To prove : BP = BQ
Proof :
AB is a common chord in both congruent circles.
In
(Sides opposite to equal of the triangle are equal )
Given :In any triangle ABC, if the angle bisector of and perpendicular bisector of BC intersect.
To prove : D lies on perpendicular bisector BC.
Construction: Join BD and DC.
Proof :
Let ABD = 1 , ADC = 2 , DCB =3 , CBD = 4
1 and 3 lies in same segment.So,
1 = 3 ..........................1(angles in same segment)
similarly, 2 = 4 ......................2
also, 1=2 ..............3(given)
From 1,2,3 , we get
3 = 4
Hence, BD = DC (angles opposite to equal sides are equal )
All points lying on perpendicular bisector BC will be equidistant from B and C.
Thus, point D also lies on perpendicular bisector BC.
Chapter No. 
Chapter Name 
Chapter 1 

Chapter 2 
CBSE NCERT solutions for class 9 maths chapter 2 Polynomials 
Chapter 3 
Solutions of NCERT class 9 maths chapter 3 Coordinate Geometry 
Chapter 4 
NCERT solutions for class 9 maths chapter 4 Linear Equations In Two Variables 
Chapter 5 
CBSE NCERT solutions for class 9 maths chapter 5 Introduction to Euclid's Geometry 
Chapter 6 

Chapter 7 

Chapter 8 
CBSE NCERT solutions for class 9 maths chapter 8 Quadrilaterals 
Chapter 9 
Solutions of NCERT class 9 maths chapter 9 Areas of Parallelograms and Triangles 
Chapter 10 
NCERT solutions for class 9 maths chapter 10 Circles 
Chapter 11 
CBSE NCERT solutions for class 9 maths chapter 11 Constructions 
Chapter 12 

Chapter 13 
NCERT solutions for class 9 maths chapter 13 Surface Area and Volumes 
Chapter 14 
CBSE NCERT solutions for class 9 maths chapter 14 Statistics 
Chapter 15 
Learn and memorize some theorems related to circles
Learn the application of those theorems in the problems
Start applying the theorems and concepts on the practice exercises.
During the practice exercises, you can take the help of NCERT solutions for class 9 maths chapter 10 Circles.
After doing all the abovewritten things, you can practice more using the previous year questions papers.
Keep working hard & happy learning!